Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, M, K, K2;
int x[10005], y[10005];
int rk[10005], rk2[10005], num[10005];
int stk[1000005][2], top;
int stk2[1000005][2], top2;
void end(int x) {
  if (!x)
    printf("-1\n");
  else {
    printf("%d\n", top + top2);
    for (int i = 1; i <= top; i++) printf("%d %d\n", stk[i][0], stk[i][1]);
    for (int i = top2; i >= 1; i--) printf("%d %d\n", stk2[i][0], stk2[i][1]);
  }
  exit(0);
}
void add(int x, int y) { stk[++top][0] = x, stk[top][1] = y; }
void add2(int x, int y) { stk2[++top2][0] = x, stk2[top2][1] = y; }
void SW(int x, int y) { add(x, y), add(y, x), add(x, y); }
void SW2(int x, int y) { add2(x, y), add2(y, x), add2(x, y); }
int main() {
  int i, j, mx, tmp;
  int fr, to;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) scanf("%d", &x[i]), rk[i] = i, rk2[i] = i;
  for (i = 1; i <= N; i++) scanf("%d", &y[i]);
  while (true) {
    mx = 0;
    for (i = K + 1; i <= N; i++)
      if (x[i] > x[mx]) mx = i;
    if (mx == 0) break;
    K++;
    swap(rk[K], rk[mx]), swap(x[K], x[mx]);
    for (i = K + 1; i <= N; i++)
      if ((x[i] ^ x[K]) < x[i]) x[i] ^= x[K], add(rk[i], rk[K]);
  }
  while (true) {
    mx = 0;
    for (i = K2 + 1; i <= N; i++)
      if (y[i] > y[mx]) mx = i;
    if (mx == 0) break;
    K2++;
    swap(rk2[K2], rk2[mx]), swap(y[K2], y[mx]);
    for (i = K2 + 1; i <= N; i++)
      if ((y[i] ^ y[K2]) < y[i]) y[i] ^= y[K2], add2(rk2[i], rk2[K2]);
  }
  if (K < K2) end(0);
  for (i = 1; i <= K2; i++) {
    tmp = x[i] ^ y[i];
    for (j = i; j <= K; j++)
      if ((tmp ^ x[j]) < tmp) tmp ^= x[j], add(rk[i], rk[j]);
    if (tmp) end(0);
  }
  for (i = K2 + 1; i <= K; i++) x[i] = 0, add(rk[i], rk[i]);
  for (i = 1; i <= N; i++) num[rk[i]] = i;
  for (i = 1; i <= K2; i++) {
    if (rk[i] != rk2[i]) {
      fr = rk[i], to = rk2[i];
      SW(fr, to);
      rk[num[to]] = fr;
      rk[i] = to;
      swap(num[fr], num[to]);
    }
  }
  end(1);
  return 0;
}